Ta có : \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1\Leftrightarrow xy+yz+zx=xyz\)

\(\sqrt{x+yz}+\sqrt{y+zx}+\sqrt{z+xy}\ge\sqrt{xyz}+\sqrt{x}+\sqrt{y}+\sqrt{z}\)

Bình phương vế trái : 

\(\left(\sqrt{x+yz}+\sqrt{y+zx}+\sqrt{z+xy}\right)^2\)

\(=\left(x+y+z+xy+yz+zx\right)+2\left(\sqrt{x+yz}.\sqrt{y+zx}+\sqrt{y+zx}.\sqrt{z+xy}+\sqrt{z+xy}.\sqrt{x+yz}\right)\)Bình phương vế phải : 

\(\left(\sqrt{xyz}+\sqrt{x}+\sqrt{y}+\sqrt{z}\right)^2=\left(xyz+x+y+z\right)+2\left(x\sqrt{yz}+y\sqrt{xz}+z\sqrt{xy}+\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\right)\)

Suy ra cần phải chứng minh : \(\sqrt{x+yz}.\sqrt{y+zx}+\sqrt{y+zx}.\sqrt{z+xy}+\sqrt{z+xy}.\sqrt{x+yz}\ge x\sqrt{yz}+y\sqrt{xz}+z\sqrt{xy}+\sqrt{x}+\sqrt{y}+\sqrt{z}\)(*)

Thật vậy, theo bđt Bunhiacopxki ta có : \(\sqrt{x+yz}.\sqrt{y+zx}\ge\sqrt{xy}+z\sqrt{xy}\)

\(\sqrt{y+zx}.\sqrt{z+xy}\ge\sqrt{yz}+x\sqrt{yz}\)

\(\sqrt{z+xy}.\sqrt{x+yz}\ge\sqrt{xz}+y\sqrt{xz}\)

Cộng các bđt trên theo vế ta chứng minh được (*) đúng.

Vậy bđt ban đầu được chứng minh.

Ý tưởng khác

Cũng từ giả thiết suy ra \(xyz=xy+yz+xz\)

Suy ra \(\sqrt{x+yz}=\sqrt{\frac{x^2+xyz}{x}}=\sqrt{\frac{x^2+xy+yz+xz}{x}}=\sqrt{\frac{\left(x+y\right)\left(x+z\right)}{x}}\)

Theo BĐT Cauchy-Schwarz ta có \(\sqrt{\left(x+y\right)\left(x+z\right)}\ge x+\sqrt{yz}\) do đó:

\(\sqrt{x+yz}=\sqrt{\frac{\left(x+y\right)\left(x+z\right)}{x}}\ge\frac{x+\sqrt{yz}}{x}=\sqrt{x}+\sqrt{\frac{yz}{x}}\)

Tương tự cho 2 BĐT còn lại \(\sqrt{y+xz}\ge\sqrt{y}+\sqrt{\frac{xz}{y}};\sqrt{z+xy}\ge\sqrt{z}+\sqrt{\frac{xy}{z}}\)

Cộng theo vế 3 BĐT được \(VT\ge\sqrt{x}+\sqrt{\frac{yz}{x}}+\sqrt{y}+\sqrt{\frac{xz}{y}}+\sqrt{z}+\sqrt{\frac{xy}{z}}\)

\(\Leftrightarrow VT\ge\sqrt{x}+\sqrt{y}+\sqrt{z}+\frac{xy+yz+xz}{\sqrt{xyz}}\)

\(\Leftrightarrow VT\ge\sqrt{x}+\sqrt{y}+\sqrt{z}+\sqrt{xyz}\) (Đpcm)

bằng 3

\(P\ge\dfrac{\sqrt{3\sqrt[3]{x^3y^3}}}{xy}+\dfrac{\sqrt{3\sqrt[3]{y^3z^3}}}{yz}+\dfrac{\sqrt{3\sqrt[3]{z^3x^3}}}{zx}\)

\(P\ge\sqrt{3}\left(\dfrac{1}{\sqrt{xy}}+\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{zx}}\right)\ge\sqrt{3}.3\sqrt[3]{\dfrac{1}{\sqrt{xy.yz.zx}}}=3\sqrt{3}\)

Dấu "=" xảy ra khi \(x=y=z=1\)

Ta có bất đẳng thức sau \(x^3+y^3\ge xy\left(x+y\right)\Leftrightarrow\left(x+y\right)\left(x-y\right)^2\ge0.\)

Do đó:

\(P=\sum\dfrac{\sqrt{1+x^3+y^3}}{xy}\ge\sum\dfrac{\sqrt{xyz+xy\left(x+y\right)}}{xy}\)

\(=\sqrt{x+y+z}\left(\dfrac{1}{\sqrt{xy}}+\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{zx}}\right)\ge\sqrt{3\sqrt[3]{xyz}}\cdot3\sqrt[3]{\dfrac{1}{\sqrt{xy}}\cdot\dfrac{1}{\sqrt{yz}}\cdot\dfrac{1}{\sqrt{zx}}}=3\sqrt{3}\)

Đẳng thức xảy ra khi $x=y=z=1.$

Áp dụng BĐT AM-GM:

\(P=\dfrac{\sqrt{1+x^3+y^3}}{xy}+\dfrac{\sqrt{1+y^3+z^3}}{yz}+\dfrac{\sqrt{1+z^3+x^3}}{zx}\)

\(\ge\dfrac{\sqrt{3xy}}{xy}+\dfrac{\sqrt{3yz}}{yz}+\dfrac{\sqrt{3zx}}{zx}\)

\(=\sqrt{3}\left(\dfrac{1}{\sqrt{xy}}+\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{zx}}\right)\)

\(\ge\sqrt{3}.3\sqrt[3]{\dfrac{1}{xyz}}=3\sqrt{3}\)

\(minP=3\sqrt{3}\Leftrightarrow x=y=z\)

Áp dụng BĐT Cô - si cho 3 số không âm:

\(1+x^3+y^3\ge3\sqrt[3]{1.x^3y^3}=3xy\Rightarrow\frac{\sqrt{1+x^3+y^3}}{xy}\ge\frac{\sqrt{3}}{\sqrt{xy}}\)

Tương tự ta có: \(\frac{\sqrt{1+y^3+z^3}}{yz}\ge\frac{\sqrt{3}}{\sqrt{yz}}\);\(\frac{\sqrt{1+z^3+x^3}}{zx}\ge\frac{\sqrt{3}}{\sqrt{zx}}\)

Cộng các vế của các BĐT trên, ta được:

\(\frac{\sqrt{1+x^3+y^3}}{xy}\)\(+\frac{\sqrt{1+y^3+z^3}}{yz}\)\(+\frac{\sqrt{1+z^3+x^3}}{zx}\ge\)\(\frac{\sqrt{3}}{\sqrt{xy}}\)\(+\frac{\sqrt{3}}{\sqrt{yz}}\)\(+\frac{\sqrt{3}}{\sqrt{zx}}\)

Tiếp tục áp dụng Cô - si:

\(BĐT\ge3\sqrt[3]{\frac{\sqrt{3}}{\sqrt{xy}}.\frac{\sqrt{3}}{\sqrt{yz}}.\frac{\sqrt{3}}{\sqrt{zx}}}=3\sqrt{3}\)

Vậy \(\frac{\sqrt{1+x^3+y^3}}{xy}\)\(+\frac{\sqrt{1+y^3+z^3}}{yz}\)\(+\frac{\sqrt{1+z^3+x^3}}{zx}\ge3\sqrt{3}\)

(Dấu "="\(\Leftrightarrow x=y=z=1\))

\(x^3+y^3+1=x^3+y^3+xyz\ge xy\left(x+y\right)+xyz=xy\left(x+y+z\right)\)

Tương tự:

\(y^3+z^3+1\ge yz\left(x+y+z\right);z^3+x^3+1\ge zx\left(x+y+z\right)\)

\(\Rightarrow VT\ge\frac{\sqrt{xy\left(x+y+z\right)}}{xy}+\frac{\sqrt{yz\left(x+y+z\right)}}{yz}+\frac{\sqrt{zx\left(x+y+z\right)}}{zx}\)

\(=\sqrt{x+y+z}\left(\frac{1}{\sqrt{xy}}+\frac{1}{\sqrt{yz}}+\frac{1}{\sqrt{zx}}\right)\)

\(\ge\sqrt{3\sqrt[3]{xyz}}\cdot3\sqrt[3]{\frac{1}{\sqrt{xy}\cdot\sqrt{yz}\cdot\sqrt{zx}}}=3\sqrt{3}\)

Dấu "=" xảy ra tại \(x=y=z=1\)

BĐT <=> \(\sqrt{\frac{x+yz}{xyz}}+\sqrt{\frac{y+xz}{xyz}}+\sqrt{\frac{z+xy}{xyz}}\ge1+\sqrt{\frac{1}{xy}}+\sqrt{\frac{1}{yz}}+\sqrt{\frac{1}{xz}}\)

Đặt \(a=\frac{1}{x};b=\frac{1}{y};c=\frac{1}{z}\)

Khi đó \(a+b+c=1\)

BĐT <=>\(\sqrt{a+bc}+\sqrt{b+ac}+\sqrt{c+ab}\ge1+\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\)

Ta có \(\sqrt{a+bc}=\sqrt{a\left(a+b+c\right)+bc}=\sqrt{\left(a+b\right)\left(a+c\right)}\ge\sqrt{\left(a+\sqrt{bc}\right)^2}=a+\sqrt{bc}\)

Khi đó \(VT\ge a+b+c+\sqrt{ab}+\sqrt{bc}+\sqrt{ac}=1+\sqrt{ab}+\sqrt{bc}+\sqrt{ac}=VP\)(ĐPCM)

Dấu bằng xảy ra khi x=y=z=3

BĐT cho tương đương với 

\(\sqrt{a+bc}+\sqrt{b+ca}+\sqrt{c+ab}\ge1+\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\)

Với \(a=\frac{1}{x};b=\frac{1}{y};c=\frac{1}{z};a+b+c=1\)

Ta có:

\(\sqrt{a+bc}=\sqrt{a\left(a+b+c\right)+bc}\)

\(=\sqrt{a^2+a\left(b+c\right)+bc}\ge\sqrt{a^2+2a\sqrt{bc}+bc}=a+\sqrt{bc}\)

Tương tự

\(\sqrt{b+ca}\ge b+\sqrt{ca};\sqrt{c+ab}\ge c+\sqrt{ab}\)

Từ đó ta có đpcm

Dấu "=" xảy ra khi x=y=z=3

rgghhjbjgtbjjjvjjjjjjgghhmbvkn  _{mm     ,,..,..guhhfxdgffdb ffd^hbvhgvg}ggg

gngn

Bất đẳng thức cần chứng minh tương đương:

\(\sqrt{x\left(x+y+z\right)+yz}+\sqrt{y\left(x+y+z\right)+zx}+\sqrt{z\left(x+y+z\right)+xy}\ge1+\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\)

\(\Leftrightarrow\sqrt{\left(x+y\right)\left(x+z\right)}+\sqrt{\left(y+z\right)\left(y+x\right)}+\sqrt{\left(z+x\right)\left(z+y\right)}\ge1+\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\). (1)

Theo bđt Bunhiakowski:

\(\sqrt{\left(x+y\right)\left(x+z\right)}\ge x+\sqrt{yz}\).

Tương tự: \(\sqrt{\left(y+z\right)\left(y+x\right)}\ge y+\sqrt{zx}\); \(\sqrt{\left(z+x\right)\left(z+y\right)}\ge z+\sqrt{xy}\).

Cộng vế với vế và kết hợp với gt x + y + z = 1 ta có (1) đúng.

Vậy ta có đpcm.

\(\sqrt{x+yz}=\sqrt{x\left(x+y+z\right)+yz}=\sqrt{\left(x+y\right)\left(x+z\right)}\ge x+\sqrt{yz}\)

Tương tự:

\(\sqrt{y+zx}\ge y+\sqrt{zx}\) ; \(\sqrt{z+xy}\ge z+\sqrt{xy}\)

Cộng vế với vế:

\(VT\ge\left(x+y+z\right)+\sqrt{xy}+\sqrt{yz}+\sqrt{zx}=...\) (đpcm)

Dấu "=" xảy ra khi \(x=y=z=\dfrac{1}{3}\)

Ta cần chứng minh:\(\dfrac{1}{\sqrt{x+y+xy}}+\dfrac{1}{\sqrt{y+z+yz}}+\dfrac{1}{\sqrt{z+x+zx}}\ge\sqrt{3}\)

Áp dụng bất đẳng thức Bunhiacopxki, ta được:

\(\dfrac{1}{\sqrt{x+y+xy}}+\dfrac{1}{\sqrt{y+z+yz}}+\dfrac{1}{\sqrt{z+x+zx}}\ge\dfrac{9}{\sqrt{x+y+xy}+\sqrt{y+z+yz}+\sqrt{z+x+zx}}\)

Mặt khác, ta có:

\(\left(\sqrt{x+y+xy}+\sqrt{y+z+yz}+\sqrt{z+x+zx}\right)^2\le3\left(\left(x+y+xy\right)+\left(y+z+yz\right)+\left(z+x+zx\right)\right)\)

\(\Leftrightarrow\left(\sqrt{x+y+xy}+\sqrt{y+z+yz}+\sqrt{z+x+zx}\right)^2\le3\left(6+xy+yz+zx\right)\)Lại có:

\(xy+yz+zx\le\dfrac{\left(x+y+z\right)^2}{3}=\dfrac{9}{3}=3\)

\(\Rightarrow\left(\sqrt{x+y+xy}+\sqrt{y+z+yz}+\sqrt{z+x+zx}\right)^2\le3\left(6+3\right)=27\)

\(\Rightarrow\sqrt{x+y+xy}+\sqrt{y+z+yz}+\sqrt{z+x+zx}\le3\sqrt{3}\)

\(\Rightarrow\dfrac{9}{\sqrt{x+y+xy}+\sqrt{y+z+yz}+\sqrt{z+x+zx}}\ge\dfrac{9}{3\sqrt{3}}=\sqrt{3}\)

Do đó \(\dfrac{1}{\sqrt{x+y+xy}}+\dfrac{1}{\sqrt{y+z+yz}}+\dfrac{1}{\sqrt{z+x+zx}}\ge\sqrt{3}\)

Dấu bằng xảy ra \(\Leftrightarrow x=y=z=1\).

Theo giả thiết: \(xyz=x+y+z+2\)

\(\Leftrightarrow xyz+xy+yz+zx+x+y+z+1\)\(=\left(xy+yz+zx\right)+2\left(x+y+z\right)+3\)

\(\Leftrightarrow\left(xy+x+y+1\right)\left(z+1\right)\)\(=\left(x+1\right)\left(y+1\right)+\left(y+1\right)\left(z+1\right)+\left(z+1\right)\left(x+1\right)\)

\(\Leftrightarrow\left(x+1\right)\left(y+1\right)\left(z+1\right)\)\(=\left(x+1\right)\left(y+1\right)+\left(y+1\right)\left(z+1\right)+\left(z+1\right)\left(x+1\right)\)

\(\Leftrightarrow\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}=1\). Đặt \(a=\frac{1}{x+1};b=\frac{1}{y+1};c=\frac{1}{z+1}\)

Khi đó a + b + c = 1 và \(x=\frac{1-a}{a}=\frac{b+c}{a}\);\(y=\frac{1-b}{b}=\frac{c+a}{b}\);\(z=\frac{1-c}{c}=\frac{a+b}{c}\)

Ta cần chứng minh \(x+y+z+6\ge2\left(\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\right)\)

\(\Leftrightarrow x+y+z+6\ge\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)^2-\left(x+y+z\right)\)

\(\Leftrightarrow\sqrt{2\left(x+y+z+3\right)}\ge\sqrt{x}+\sqrt{y}+\sqrt{z}\)

\(\Leftrightarrow\sqrt{2\left[\left(x+1\right)+\left(y+1\right)+\left(z+1\right)\right]}\ge\sqrt{x}+\sqrt{y}+\sqrt{z}\)

\(\Leftrightarrow\sqrt{\left[\left(b+c\right)+\left(c+a\right)+\left(a+b\right)\right]\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)}\)\(\ge\sqrt{\frac{b+c}{a}}+\sqrt{\frac{c+a}{b}}+\sqrt{\frac{a+b}{c}}\)

BĐT cuối hiển nhiên đúng vì đây là BĐT Bunyakovski do đó bài toán được chứng minh.

Đẳng thức xảy ra khi \(a=b=c=\frac{1}{3}\)hay x = y = z = 2